Answer :
To find the answers, we will follow a step-by-step approach using principles from electromagnetism.
### Part (a): Finding the electric flux through the Gaussian surface
1. Identify the given charges:
- [tex]\( q_1 = 2.73 \ \mu \text{C} \)[/tex]
- [tex]\( q_2 = -3.98 \ \mu \text{C} \)[/tex]
- [tex]\( q_3 = 5.83 \ \mu \text{C} \)[/tex]
2. Determine the net charge enclosed by the Gaussian surface:
The net charge enclosed [tex]\( (Q_{\text{net}}) \)[/tex] is the sum of all the charges inside the Gaussian surface:
[tex]\[ Q_{\text{net}} = q_1 + q_2 + q_3 \][/tex]
Substituting in the given values, we get:
[tex]\[ Q_{\text{net}} = 2.73 \ \mu \text{C} + (-3.98 \ \mu \text{C}) + 5.83 \ \mu \text{C} \][/tex]
[tex]\[ Q_{\text{net}} = 2.73 - 3.98 + 5.83 = 4.58 \ \mu \text{C} \][/tex]
3. Apply Gauss's Law:
According to Gauss's Law, the electric flux [tex]\( \Phi \)[/tex] through a closed surface is given by:
[tex]\[ \Phi = \frac{Q_{\text{net}}}{\epsilon_0} \][/tex]
where [tex]\( \epsilon_0 \)[/tex] is the permittivity of free space.
4. Calculate the electric flux:
Since [tex]\( Q_{\text{net}} \)[/tex] is [tex]\( 4.58 \ \mu \text{C} \)[/tex], the electric flux [tex]\( \Phi \)[/tex] is determined by this charge divided by [tex]\( \epsilon_0 \)[/tex]. However, in this problem, we are not asked to calculate the exact value of the flux. Instead, we leave it as a function of [tex]\( Q_{\text{net}} \)[/tex].
Thus,
[tex]\[ \boxed{\Phi = \frac{4.58 \times 10^{-6} \ \text{C}}{\epsilon_0}} \][/tex]
### Part (b): Finding the required charge [tex]\( Q \)[/tex] to reduce the electric flux to zero
1. Determine the condition for zero flux:
To reduce the electric flux through the surface to zero, the net charge enclosed by the surface must be zero. This means the sum of all enclosed charges must be zero.
2. Set up the equation for zero net charge:
The sum of the original charges [tex]\( Q_{\text{net}} \)[/tex] and the new charge [tex]\( Q \)[/tex] must be zero:
[tex]\[ Q_{\text{net}} + Q = 0 \][/tex]
Given [tex]\( Q_{\text{net}} = 4.58 \ \mu \text{C} \)[/tex]:
[tex]\[ 4.58 \ \mu \text{C} + Q = 0 \][/tex]
3. Solve for [tex]\( Q \)[/tex]:
[tex]\[ Q = -4.58 \ \mu \text{C} \][/tex]
4. Determine the sign and magnitude of [tex]\( Q \)[/tex]:
The magnitude of [tex]\( Q \)[/tex] is [tex]\( 4.58 \ \mu \text{C} \)[/tex] and the sign is negative, as [tex]\( Q \)[/tex] must offset the positive net charge [tex]\( Q_{\text{net}} \)[/tex].
Therefore, the charge [tex]\( Q \)[/tex] required to reduce the electric flux through the surface to zero is:
[tex]\[ \boxed{Q = -4.58 \ \mu \text{C}} \][/tex]
### Part (a): Finding the electric flux through the Gaussian surface
1. Identify the given charges:
- [tex]\( q_1 = 2.73 \ \mu \text{C} \)[/tex]
- [tex]\( q_2 = -3.98 \ \mu \text{C} \)[/tex]
- [tex]\( q_3 = 5.83 \ \mu \text{C} \)[/tex]
2. Determine the net charge enclosed by the Gaussian surface:
The net charge enclosed [tex]\( (Q_{\text{net}}) \)[/tex] is the sum of all the charges inside the Gaussian surface:
[tex]\[ Q_{\text{net}} = q_1 + q_2 + q_3 \][/tex]
Substituting in the given values, we get:
[tex]\[ Q_{\text{net}} = 2.73 \ \mu \text{C} + (-3.98 \ \mu \text{C}) + 5.83 \ \mu \text{C} \][/tex]
[tex]\[ Q_{\text{net}} = 2.73 - 3.98 + 5.83 = 4.58 \ \mu \text{C} \][/tex]
3. Apply Gauss's Law:
According to Gauss's Law, the electric flux [tex]\( \Phi \)[/tex] through a closed surface is given by:
[tex]\[ \Phi = \frac{Q_{\text{net}}}{\epsilon_0} \][/tex]
where [tex]\( \epsilon_0 \)[/tex] is the permittivity of free space.
4. Calculate the electric flux:
Since [tex]\( Q_{\text{net}} \)[/tex] is [tex]\( 4.58 \ \mu \text{C} \)[/tex], the electric flux [tex]\( \Phi \)[/tex] is determined by this charge divided by [tex]\( \epsilon_0 \)[/tex]. However, in this problem, we are not asked to calculate the exact value of the flux. Instead, we leave it as a function of [tex]\( Q_{\text{net}} \)[/tex].
Thus,
[tex]\[ \boxed{\Phi = \frac{4.58 \times 10^{-6} \ \text{C}}{\epsilon_0}} \][/tex]
### Part (b): Finding the required charge [tex]\( Q \)[/tex] to reduce the electric flux to zero
1. Determine the condition for zero flux:
To reduce the electric flux through the surface to zero, the net charge enclosed by the surface must be zero. This means the sum of all enclosed charges must be zero.
2. Set up the equation for zero net charge:
The sum of the original charges [tex]\( Q_{\text{net}} \)[/tex] and the new charge [tex]\( Q \)[/tex] must be zero:
[tex]\[ Q_{\text{net}} + Q = 0 \][/tex]
Given [tex]\( Q_{\text{net}} = 4.58 \ \mu \text{C} \)[/tex]:
[tex]\[ 4.58 \ \mu \text{C} + Q = 0 \][/tex]
3. Solve for [tex]\( Q \)[/tex]:
[tex]\[ Q = -4.58 \ \mu \text{C} \][/tex]
4. Determine the sign and magnitude of [tex]\( Q \)[/tex]:
The magnitude of [tex]\( Q \)[/tex] is [tex]\( 4.58 \ \mu \text{C} \)[/tex] and the sign is negative, as [tex]\( Q \)[/tex] must offset the positive net charge [tex]\( Q_{\text{net}} \)[/tex].
Therefore, the charge [tex]\( Q \)[/tex] required to reduce the electric flux through the surface to zero is:
[tex]\[ \boxed{Q = -4.58 \ \mu \text{C}} \][/tex]
